Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration by substitution (and esp. Weierstrass' substitution)

  1. Feb 24, 2012 #1
    The stupid question of the day.

    Is it fair to say that[tex]\frac{du}{dx} = \frac 1 {dx/du}
    [/tex]since this comes (I think) from the chain rule,[tex]
    \frac{dx}{du} \frac{du}{dx} = \frac{dx}{dx} = 1

    Which means that, when integrating by substitution, I can choose to do either of[tex]
    \int f(u) du = \int f(x) \frac {du}{dx} dx = \int \frac {f(x)} {dx/du} dx
    [/tex]depending on which derivative I happen to have at hand.

    (Just checking; you can't be too careful when treating differentials as it they were fractions, which they aren't.)

    The matter came out while studying Weierstrass' substitution, where[tex]
    \int \frac 1 {\sin x} dx &= \int \frac {\frac{1+t^2}{2t}} {\frac{1+t^2}{2}}
    = \int \frac 1 t dt &\mbox{with }t=\tan \frac x 2
    [/tex]because [itex]\frac {dt}{dx}[/itex] is easier to figure out than [itex]\frac {dx}{dt}[/itex].

    Any more detail (particularly from the viewpoint of analysis) is welcome. Right now I feel like I'm just shuffling symbols without really knowing what am I doing.
  2. jcsd
  3. Feb 24, 2012 #2


    User Avatar
    Science Advisor

    As long as u(x) is invertible (i.e. as long as du/dx is not 0), yes, [itex]dx/du= 1/(du/dx)[/itex].

    Your second statement is also correct but you have to be careful with it. Unless du/dx is a constant we cannot just shift du/dx inside or outside the integral. But we can use one or the other of those forms, whichever is appropriate.
  4. Feb 24, 2012 #3
    Hey, Halls, thanks.
    Yes, I absolutely agree! The intention was, as you say next, choosing one form or the other, depending on which derivative I find easier to get.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook